Integrate. $ \int 9\sin(x)\,dx $ $=$ $+ C$
Answer: We need a function whose derivative is $9\sin(x)$. We know that the derivative of $\cos(x)$ is $-\sin(x)$, so let's start there: $\dfrac{d}{dx} \cos(x) = -\sin(x)$ Now let's add in a negative sign: $\dfrac{d}{dx}\left[ -\cos(x)\right] = \sin(x)$ Now let's multiply by $9$ : $\dfrac{d}{dx}\left[ -9\cos(x) \right]= 9\sin(x)$ Because finding the integral is the opposite of taking the derivative, this means that: $ \int 9\sin(x)\,dx =-9 \cos(x)\, + C$ The answer: $-9 \cos(x)\, + C$